The tangent function, denoted \(\tan(x)\), is one of the six common trigonometric functions. Tangent is most commonly related to sine and cosine as
$$\tan(x) = \frac{\sin(x)}{\cos(x)}$$
This will be central to many manipulations with expressions we use involving tangent. Since the tangent function is defined as it is, there are values of \(x\) such that the tangent function is undefined, and this problem is fundamental to graphical properties of the tangent function.
Example 1: Find all values of \(x\) on the interval \([0, 2\pi]\) such that \(\tan(x)\) is undefined.
Solution: We use the definition of tangent to rewrite it as
$$\frac{\sin(x)}{\cos(x)}$$
The fraction is undefined where the denominator is \(0\), so we wish to solve the equation
$$\cos(x) = 0$$
In the given domain, the solutions are \(x = \frac{\pi}{2}\) and \(x = \frac{3\pi}{2}\), according to the arccosine function.
This is important to know when doing certain calculations.
One of the most common identities in trigonometry is called Pythagoras's Identity, which states that for all \(x\):
$$\sin^2 (x) + \cos^2(x) = 1$$
But what does this have to do with tangent? We can derive an identity with this identity that involves tangent.
Example 2: Consider Pythagoras's Identity:
$$\sin^2 (x) + \cos^2(x) = 1$$
To introduce a tangent function into the equation, recall that
$$\frac{\sin(x)}{\cos(x)} = \tan(x)$$
We can square both sides of this equation:
$$\frac{\sin^2(x)}{\cos^2(x)} = \tan^2(x)$$
Therefore, dividing Pythagoras's Identity by \(\cos^2(x)\) introduces a tangent function:
$$\frac{\sin^2(x)}{\cos^2(x)} + \frac{\cos^2(x)}{\cos^2(x)} = \frac{1}{\cos^2(x)} \Rightarrow$$ $$\tan^2(x) + 1 = \sec^2(x)$$
This identity is very important for some expressions involving tangent:
$$\tan^2(x) + 1 = \sec^2(x)$$
Let's see how this identity can be applied.
Example 3: Simplify the expression \(\sec^2(x)(\tan^2(x) - \sin^2(x) - \cos^2(x))\), writing it in terms of tangent functions.
Solution: The second factor is more obvious in terms of how to simplify it. Rewrite the expression as
$$\sec^2(x)(\tan^2(x) - (\sin^2(x) + \cos^2(x)))$$
This manipulation is helpful because \(\sin^2(x) + \cos^2(x) = 1\), according to Pythagoras's Identity:
$$\sec^2(x)(\tan^2(x) - 1)$$
We know by our other identity that \(\sec^2(x) = \tan^2(x) + 1\), so we can substitute this into our expression to get rid of all the trigonometric functions except for tangent:
$$(\tan^2(x) + 1)(\tan^2(x) - 1)$$
Finally, we can expand the product, which is quite easy since it is a difference of squares:
$$(\tan^2(x) + 1)(\tan^2(x) - 1) = \tan^4(x) - 1$$
That is just one example of how the trigonometric identities can be applied. Here is one more:
Example 4: Solve this equation in the domain \([0, 2\pi]\):
$$\frac{\sin^4(x)}{1 - \sin^2(x)} - \sin^2(x) - 2\sec^2(x) + 4 = 0$$
Solution: Your goal should be to rewrite the equation in terms of as few trigonometric functions as possible. Since Pythagoras's Identity states that
$$\sin^2(x) + \cos^2(x) = 1$$
we can subtract \(\sin^2(x)\) from both sides of the equation to get
$$\cos^2(x) = 1 - \sin^2(x)$$
Substitute this into the first term of the left-hand side of the original equation:
$$\frac{\sin^4(x)}{\cos^2(x)} - \sin^2(x) - 2\sec^2(x) + 4 = 0$$
As a result, we are well on our way of writing this equation only with sine and tangent. The first term can be simplified by remembering that \(\frac{\sin(x)}{\cos(x)} = \tan(x)\):
$$\frac{\sin^4(x)}{\cos^2(x)} = \sin^2(x) \cdot \frac{\sin^2(x)}{\cos^2(x)} = \sin^2(x) \cdot (\frac{\sin(x)}{\cos(x)})^2 = \sin^2(x)\tan^2(x)$$
Thus the equation becomes
$$\sin^2(x)\tan^2(x) - \sin^2(x) - 2\sec^2(x) + 4 = 0$$
In a similar manner as before, since
$$1 + \tan^2(x) = \sec^2(x)$$
we can substitute the secant in for an expression only involving tangent:
$$\sin^2(x)\tan^2(x) - \sin^2(x) - 2(1 + \tan^2(x)) + 4 = 0$$
Simplify:
$$\sin^2(x)\tan^2(x) - \sin^2(x) - 2 - 2\tan^2(x) + 4 = 0 \Rightarrow$$ $$\sin^2(x)\tan^2(x) - \sin^2(x) - 2\tan^2(x) + 2 = 0$$
Converting the equation to only sine and tangent lets us factor it:
$$\sin^2(x)(\tan^2(x) - 1) - 2(\tan^2(x) - 1) = 0 \Rightarrow$$ $$(\sin^2(x) - 2)(\tan^2(x) - 1) = 0$$
The factor \(\sin^2(x) - 2\) can be equated to \(0\) to find the relevant solutions:
$$\sin^2(x) - 2 = 0 \Rightarrow$$ $$\sin^2(x) = 2 \Rightarrow$$ $$\sin(x) = \pm \sqrt{2}$$
However, the range of \(\sin(x)\) is \([-1, 1]\), and both possible values of \(\sin(x)\) are outside this range, so this factor of the equation yields no real solutions.
Therefore we equate the other factor to \(0\):
$$\tan^2(x) - 1 = 0$$
Solve for \(\tan(x)\):
$$\tan^2(x) = 1 \Rightarrow$$ $$\tan(x) = \pm 1$$
In the domain we are using, considering the graph of tangent gives us \(x = \frac{\pi}{4}\) and \(x = \frac{5\pi}{4}\) when \(\tan(x) = 1\) and \(x = \frac{3\pi}{4}\) and \(x = \frac{7\pi}{4}\) when \(\tan(x) = -1\).
The tangent function has a derivative that is sometimes useful:
$$\frac{d}{dx}(\tan(x)) = \sec^2(x)$$
This can be helpful in conjunction with the trigonometric identities we showed earlier to make many differentiation calculations easier.
Example 5: Find the derivative of \(\sec^2(x) - 1\).
Solution: What does this have to do with tangent? Well, since
$$\tan^2(x) + 1 = \sec^2(x)$$
we can solve for \(\tan^2(x)\):
$$\tan^2(x) = \sec^2(x) - 1$$
Therefore we wish to differentiate:
$$\tan^2(x)$$
This is easy with our derivative of tangent and the Chain Rule for differentiation. For clarity, the inside function is \(\tan(x)\) and the outside function is \(x^2\):
$$\frac{d}{dx}(\tan^2(x)) = 2\tan(x)\frac{d}{dx}(\tan(x)) = 2\tan(x)\sec^2(x)$$
Example 6: Find the derivative of \(\tan(x)\sin(2x)\).
Solution: We must use the Product Rule and the Chain Rule, recalling that \(\frac{d}{dx}(\sin(x)) = \cos(x)\):
$$\frac{d}{dx}(\tan(x)\sin(2x)) = \sin(2x)\frac{d}{dx}(\tan(x)) + \tan(x)\frac{d}{dx}(\sin(2x)) = \sin(2x)\sec^2(x) + 2\cos(2x)\tan(x)$$
Since
$$\frac{d}{dx}(\tan(x)) = \sec^2(x)$$
We can integrate this equation to get
$$\int{\sec^2(x)dx} = \tan(x) + C$$
This is helpful for various integration problems.
Example 7: Integrate \(\int{\frac{\sin^2(x) + 2\cos^2(x)}{\cos^2(x)}dx}\).
Solution: Break up the fraction using the definition of tangent:
$$\int{\frac{\sin^2(x) + 2\cos^2(x)}{\cos^2(x)}dx} = \int{\frac{\sin^2(x)}{\cos^2(x)} + \frac{2\cos^2(x)}{\cos^2(x)}dx} = \int{\tan^2(x) + 2}dx$$
It is easier to integrate squares of secants than squares of tangents, so convert the tangent to a secant:
$$\int{\tan^2(x) + 2}dx = \int{\tan^2(x) + 1 + 1}dx = \int{\sec^2(x) + 1}$$
Therefore we can use the integration formula above:
$$\int{\sec^2(x) + 1}dx = \int{\sec^2(x)dx} + \int{1dx} = \tan(x) + x + C$$
Example 8: Integrate \(\int{\frac{(1 - \cos(x))(1 + \cos(x))}{(1 + \sin(x))(1 - \sin(x))}dx}\).
Solution: The numerator and denominator are both differences of squares, so expand the products:
$$\int{\frac{1 - \cos^2(x)}{1 - \sin^2(x)}dx}$$
We know that from Pythagoras's Identity, we can isolate the sine and cosine terms to derive
$$1 - \sin^2(x) = \cos^2(x)$$ $$1 - \cos^2(x) = \sin^2(x)$$
These are important identities and should be remembered. Therefore the integrand becomes
$$\int{\frac{\sin^2(x)}{\cos^2(x)}dx}$$
This is equivalent to
$$\int{\tan^2(x)dx}$$
We can convert this to a secant by rewriting it as
$$\int{\tan^2(x) + 1 - 1dx}$$
This is equal to
$$\int{\sec^2(x) - 1}dx$$
Now integrate:
$$\tan(x) - x + C$$
But how do you integrate a lone tangent function? We use a technique called u-substitution, where we substitute a function \(u = f(x)\) and use it and \(\frac{du}{dx}\) to simplify the expression.
Example 9: Integrate \(\int{\tan(x)}dx\).
Solution: We rewrite the integrand as a quotient:
$$\int{\frac{\sin(x)}{\cos(x)}dx}$$
We can create a u-substitution here. Let \(u = \cos(x)\). Then the integral becomes
$$\int{\frac{\sin(x)}{u}dx}$$
This substitution is helpful because
$$\frac{du}{dx} = -\sin(x)$$
Therefore the expression \(-\sin(x)\) can be substituted out:
$$-\int{\frac{\frac{du}{dx}}{u}dx} = -\int{\frac{1}{u}du}$$
This integral is easy to calculate--it is the opposite of the natural logarithm:
$$-\ln|u| + C$$
Now undo the substitution:
$$-\ln|\cos(x)| + C$$
If we wanted, we could rewrite the logarithm to remove the negative sign:
$$\ln|\sec(x)| + C$$
This integral is also very important:
$$\int{\tan(x)dx} = \ln|\sec(x)| + C$$
Let's see how this can be applied to an integration problem.
Example 10: Integrate \(\int{\sin(x)(\sec(x) + \cos(x))dx}\).
Solution: We can distribute the left factor:
$$\int{\sin(x)\sec(x) + \sin(x)\cos(x)dx}$$
Both of these terms are fairly easy to integrate. Starting with the first one, realize that \(\sec(x) = \frac{1}{\cos(x)}\) and so the integral becomes
$$\int{\frac{\sin(x)}{\cos(x)} + \sin(x)\cos(x)dx} = \int{\tan(x) + \sin(x)\cos(x)dx}$$
Break up the integral into two pieces so we can integrate the first one:
$$\int{\tan(x)dx} + \int{\sin(x)\cos(x)dx} = \ln|\sec(x)| + \int{\sin(x)\cos(x)dx}$$
The second one is interesting, because we can apply the double-angle identity for sine, \(2\sin(x)\cos(x) = \sin(2x)\), making it easier to integrate. Introduce the necessary \(2\) into the integrand by multiplying the remaining integral by \(\frac{1}{2} \cdot 2\):
$$\ln|\sec(x)| + \frac{1}{2}\int{2\sin(x)\cos(x)dx} = \ln|\sec(x)| + \frac{1}{2}\int{\sin(2x)dx}$$
Now we can perform the Chain Rule backwards on this integral, remembering that the antiderivative of \(\sin(x)\) is \(-\cos(x)\):
$$\ln|\sec(x)| - \frac{1}{4}\cos(2x) + C$$
The tangent function, as we can see, is very useful for a variety of problems in which sine and cosine, the more common trigonometric functions, will not cut it alone. Being able to apply the basic concepts and identities will be very helpful to you as you continue to solve problems.